082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

25.6 Taylor polynomials and Taylor series in two variables 839
so
∂p 2
∂x (2,−1)=2+2(−1)+4=4 ∂p 2
∂y (2,−1)=1+4−8=−3
Hence
∂p 2
∂f
(2,−1) =
∂x ∂x (2,−1) ∂p 2
∂f
(2,−1) =
∂y ∂y (2,−1)
that is, the first partial derivatives of the Taylor polynomial and the function have
identical values at (2,−1).
(e) The second partialderivatives of p 2
arefound:
∂ 2 p 2
∂x 2 = 1
Example25.13 Afunction f isgiven by
∂ 2 p 2
∂x∂y = 2 ∂ 2 p 2
∂y 2 = −1
These values areidentical tothe second partial derivatives of f at (2,−1).
f(x,y)=x 3 +x 2 y+y 4
(a) State the second-order Taylor polynomialgeneratedby f about(1, 1).
(b) Usethepolynomialtoestimate f(1.2,0.9).Comparethisvaluewiththetruevalue.
(c) Verify that the second partial derivatives of the function and the Taylor polynomial
are identical at(1,1).
Solution (a) Herea = 1 andb= 1.We aregiven that f =x 3 +x 2 y +y 4 and so
∂f
∂x = 3x2 +2xy
∂ 2 f
∂x∂y = 2x
∂f
∂y =x2 +4y 3
∂ 2 f
∂y 2 = 12y2
Evaluation of f and itsderivatives at(1, 1) yields
f = 3
∂f
∂x = 5
∂f
∂y = 5
∂ 2 f
∂x 2 = 8
The second-order Taylor polynomial is p 2
(x,y):
∂ 2 f
∂x 2 =6x+2y
∂ 2 f
∂x∂y = 2
p 2
(x,y)=f +(x−1) ∂f
∂x +(y−1)∂f ∂y
+ 1 (
)
(x−1) 2∂2 f
2 ∂x +2(x −1)(y −1) ∂2 f f
2 ∂x∂y +(y−1)2∂2 ∂y 2
∂ 2 f
∂y 2 = 12
=3+5(x−1)+5(y−1)+ 1 2 (8(x −1)2 +4(x −1)(y −1) +12(y −1) 2 )
=4x 2 +6y 2 +2xy−5x−9y+5
(b) The value of p 2
(1.2,0.9) isanestimate of f(1.2,0.9).
p 2
(1.2,0.9) = 4(1.2) 2 +6(0.9) 2 +2(1.2)(0.9) −5(1.2) −9(0.9) +5 = 3.68
The actual value is
f(1.2,0.9) = (1.2) 3 + (1.2) 2 (0.9) + (0.9) 4 = 3.6801